Đáp án: $P > 3$
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:x > 0;x\# 1\\
P = 1:\left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{{\sqrt x + 1}}{{x - 1}}} \right)\\
= 1:\left( {\dfrac{{x + 2}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} + \dfrac{{\sqrt x + 1}}{{x + \sqrt x + 1}} - \dfrac{1}{{\sqrt x - 1}}} \right)\\
= 1:\dfrac{{x + 2 + \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x + 2 + x - 1 - x - \sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{x - \sqrt x }}\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }}\\
P - 3\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x }} - 3\\
= \dfrac{{x + \sqrt x + 1 - 3\sqrt x }}{{\sqrt x }}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} > 0\left( {khi:x > 0;x\# 1} \right)\\
\Leftrightarrow P > 3
\end{array}$
